= Solution
At this separation the <orbital overlap> and interatomic <two-electron exchange integrals> are negligible. Each localized <electron> then contributes essentially the isolated <hydrogen> energy, while the interatomic electrostatic terms cancel to leading order for neutral fragments. Including nuclear repulsion in the molecular energy gives
$$
\boxed{E_{\rm UHF}\simeq2(-\tfrac12 E_h)=-1E_h\simeq-27.2\ \mathrm{eV}.}
$$
The closed-shell <RHF> <determinant> instead doubly occupies $g\simeq(A+B)/\sqrt2$. It retains ionic configurations with both <electrons> on the same atom, even at dissociation. Using <atomic units> and the <Hartree energy> in the stated unrelaxed hydrogenic 1s basis, the same-atom <two-electron Coulomb integral> is $J_{AA}=J_{BB}=5E_h/8$. Consequently at infinite separation
$$
E_{\rm RHF}\simeq-1+\tfrac14(J_{AA}+J_{BB})=-\frac{11}{16}E_h\simeq-18.7\ \mathrm{eV},
$$
where energies and integrals inside the first expression are in <atomic units>. It is much too high compared with two neutral atoms. Retaining the leading intersite Coulomb term at $R=10\ \text{Å}=18.90a_0$ gives $E_{\rm RHF}\simeq-11/16-1/(2R)\simeq-0.714E_h$, still about $0.286E_h$ above <UHF>. The numerical <RHF> estimates specify the supplied fixed 1s basis; a larger optimized orbital space changes them. If nuclear repulsion is excluded by an electronic-energy convention, subtract the same $1/R$ from both quoted total molecular energies. The restricted/unrestricted comparison and their dissociation behaviour are unchanged.<XXXIV. Notes on the molecular orbital treatment of the hydrogen molecule by Charles Alfred Coulson and Inga Fischer-Hjalmars>[1949, pp. 386–393, minimal atomic-1s molecular-orbital treatment]{ref}
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